Genetics: Traits in F2 generation Punnette Square

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

Dr. Freedom

New Member
7+ Year Member
Joined
Aug 23, 2014
Messages
1
Reaction score
0
Hello All! This is a fairly basic genetics question but I was wondering if someone would be willing to take me through the thought process involved. I've always struggled with these types of questions. I would think two Punnett Square's would be involved for the F1 and F2 generations, but I'm not sure how to set them up. In your answers, I would appreciate if it could be specified what each part of the question is trying to imply and paint the whole picture for me in arriving to the answer to your question. Thank you!

A guppy that is homozygous for the traits small size and late maturation (the dominant traits), mates with a guppy that is large and early-maturing, producing F1 offspring. If two of the F1 guppies mate, what percentage of the F2 offspring is expected to be homozygous for those traits favorable to life in the tributary? [Note: Assume all traits are on autosomal chromosomes.]

A. 6.25%
B. 12.5%
C. 25%
D. 50%

For context about the passage see the attached screenshot jpeg.

Screen Shot 2017-01-08 at 11.24.56 AM.jpg

Members don't see this ad.
 
Last edited:
In the F1 generation, they all have to be heterozygous because if a homozygous dominant mates with a homozygous recessive (which must be the case for parent 2), then the offspring will all be heterozygous for the trait and all express the dominant phenotype. Now, the question is reduced to: what happens when two individuals heterozygous for the size and maturation traits mate? Well, you then need to use a Punnett square to figure out all the combinations. Let's assume that S = small, s = large, L = late maturation, and l = early maturation. The mating individuals then are SsLl and SsLl. Thus, each parent can pass on each of the following combinations: SL, Sl, sL, and sl. Put these in a Punnett square that's 4x4 and you get the following combinations: SSLL, SSLl x 2, SsLL x 2, SsLl x 4, SSll, Ssll x 2, ssLL, ssLl x 2, ssll.

Now, you want large, late-maturing offspring that are homozygous for these traits. That is, SSLL, only one of those sixteen combinations will give you that. So either the answer is A, or I did the cross wrong somewhere, which is pretty likely. But the idea is there.

You can also come to this conclusion by just making two 2x2 Punnett squares since the traits are independent of each other. Crossing an Ss with another Ss results in 1/4 of the offspring being SS. Same with Ll with Ll. Therefore, for both SS and LL to be in the same offspring, the chances are 1/4*1/4, or 1/16.
 
Top